# How many solutions has the equation 4^x-2^x=56?

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The equation 4^x-2^x=56 can be considered to be a quadratic equation of the form ay^2 + by + c = 0 with y = 2^x

Solving the equation with the given substitution:

y^2 - y = 56

y^2 - y - 56 = 0

y^2 - 8y + 7y - 56 = 0

y(x - 8) + 7(y - 8) = 0

(y + 7)(y - 8) = 0

y = -7 and y = 8

Now, the power of a positive number can never be negative.

As y = 2^x and 2 is positive y cannot be equal to -7. The only solution of 2^x is 8.

2^x = 8 = 2^3

x = 3

The given equation has only one solution.

We'll create a common base, so we'll write 4^x = 2^2x.

We'll re-write th equation:

2^2x - 2^x - 56 = 0

We'll replace 2^x by t:

t^2 - t - 56 = 0

We'll apply quadratic formula:

t1 = [1 + sqrt(1+224)]/2

t1 = (1 + sqrt225)/2

t1 = (1+15)/2

t1 = 8

t2 = (1-15)/2

t2 = -7

We'll put 2^x = t1 => 2^x = 8

We'll create matching bases:

2^x = 2^3

Since the bases are matching, we'll apply one to one property:

x = 3

We'll put 2^x = t2 => 2^x = -7 impossible.

**The equation will have only one solution and this one is x = 3.**